# Find the infinite sum of $\sum_{n=0}^\infty \ \frac{1}{(2n)!}$ and $\sum_{n=0}^\infty \ \frac{n}{2^n}$.

I need help with this problem:

Find the following infinite sums. (Most of the cases are equal to $$f(a)$$ where $$a$$ is an obvious number and $$f(x)$$ is defined by a power series. To calculate the series, it is necessary to effect the necessary arrangements until there appear some well-known power series.)

1. $$\sum_{n=0}^\infty \ \frac{1}{(2n)!}$$
2. $$\sum_{n=0}^\infty \ \frac{n}{2^n}$$

For the first one, I really don't know how to begin. For the second one, I started by rewriting it like: $$\sum_{n=0}^\infty \ n(\frac{1}{2})^n$$ so it looks like a geometric series; after that I don't know what to do. Is it ok if I divide it by n and end up with $$\sum_{n=0}^\infty \ (\frac{1}{2})^n=\frac{1}{1-\frac{1}{2}}=2$$?

Note that $$\cosh(z)=\sum_{n=0}^{\infty}{\frac{1}{(2n)!}z^{2n}}$$. So, $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}=\cosh(1)$$

For the second, if $$|z|<1$$, then $$\sum_{n=0}^{\infty}{z^n}=\dfrac{1}{1-z}$$, I suppose you know that. Therefore, $$\sum_{n=1}^{\infty}{nz^{n-1}}=\dfrac{1}{(1-z)^2}\implies \sum_{n=0}^{\infty}{nz^{n}}=\dfrac{z}{(1-z)^2}$$ Now, take $$z=\frac{1}{2}.$$

I will explain the first sum using exponential. We have $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}$$, each time you see factorial is a good idea think in exponential series. So, note that in this series you have only "even factorials", So I'll try to form this series. Note that $$e^{x}+e^{-x}=\sum_{n=0}^{\infty}{\frac{1}{n!}x^{n}}+\sum_{n=0}^{\infty}{\frac{1}{n!}(-1)^{n}x^{n}}=\sum_{n=0}^{\infty}{\frac{1}{n!}x^{n}(1+(-1)^n)}$$

Clearly, the terms $$1+(-1)^n=\{0,2\}$$ if $$n$$ is even or odd number. so, we only considered the even numbers, i.e., $$e^{x}+e^{-x}=\sum_{n=0}^{\infty}{\frac{1}{n!}x^{n}(1+(-1)^n)}=\sum_{n=0}^{\infty}{\frac{2}{(2n)!}x^{n}}$$

You can see that this looks a lot like the series we are looking for. Now take $$x=1$$ in the above,

$$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}=\frac{1}{2}(e^{1}+e^{-1})$$

And using the complex form, $$e^{ix}=\cos(x)+i\sin(x)$$ where $$i=\sqrt{-1}$$, you can see $$e^{1}+e^{-1}=e^{i(-i)}+e^{i(i)}=(\cos(-i)+i\sin(-i))+(\cos(i)+i\sin(i))$$ $$=(\cos(i)-i\sin(i))+(\cos(i)+i\sin(1))=2\cos(i)$$ And we "define", $$\cos(i)=\cosh(1)$$. Finally, $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}=\frac{1}{2}(e^{1}+e^{-1})=\cos(i)=\cosh(1)$$

• How do you know that? I had no idea that $\cosh(x)$ was equal to that. I think that I'm not suppose to now that, because that series for $\cosh(x)$ isn't anywhere in my book. – davidllerenav Feb 11 at 4:32
• I saw the answer on the book, and it used $\e^x$ and $\e^-x$, but I don't understand it. – davidllerenav Feb 11 at 4:33
• @davidllerenav use the exponential form of the $\cosh$ function. – aleden Feb 11 at 4:42
• cosh($x$)=$(e^x+e^{-x})/2$ – J. W. Tanner Feb 11 at 4:49
• In general we can write $$\sum_{n=0}^N a_{2n}=\frac12\sum_{n=0}^{2N} (1+(-1)^n)a_n$$Setting $a_n=\frac1{n!}$ and letting $N\to \infty$, we find that $$\sum_{n=0}^\infty \frac1{(2n)!}=\frac12\left(\sum_{n=0}^\infty \frac1{n!}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}\right)=\frac12(e+e^{-1})$$ – Mark Viola Feb 11 at 6:01

For the second sum, $$f(x)=\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ $$f'(x)=\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ $$\frac{x}{(1-x)^2}=\sum_{n=0}^\infty nx^n$$ For $$x=\frac{1}{2}$$, $$\frac{\frac{1}{2}}{(1-\frac{1}{2})^2}=2=\sum_{n=0}^\infty \frac{n}{2^n}$$

• You multiplied $f'(x)=\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$ by x, right? That's why we end up with $\frac{x}{(1-x)^2}=\sum_{n=0}^\infty nx^n$? – davidllerenav Feb 11 at 4:50
• @davidllerenav Yep, the crucial step is taking the derivative in order to get the $n$ term as a coefficient in the power series. – aleden Feb 11 at 4:59
• Ok, thanks. I understand that. – davidllerenav Feb 11 at 5:36

For the first problem : $$\cos x = 1- \frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}......$$ Put the value of $$x$$ as $$i$$ to get: $$\cos (ix) = 1- \frac{(ix)^2}{2!} + \frac{(ix)^4}{4!}-\frac{(ix)^6}{6!}......$$ Solve further to get: $$\cos (ix) = 1+ \frac{x^2}{2!} + \frac{x^4}{4!}+\frac{x^6}{6!}......$$ Put the value of $$x=1$$ to get $$\cos i = 1+ \frac{1}{2!} + \frac{1}{4!}+\frac{1}{6!}......=\sum_{n=0}^\infty \ \frac{1}{(2n)!}$$ The second series is an arithmetico geometric series with the $$n^{th}$$ term as: $$T_n=\frac{n}{2^n}$$ when you calculate the sum(start the sum from n=1 as it wont matter): $$S_{\infty}= \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}.....$$ You can see that the numerators are in AP where as the fractions are in GP. Just multiply the above expression with the common ratio i.e. $$\frac{1}{2}$$ to get $$\frac{1}{2}S_{\infty}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}...$$ Subtract both of these equations to get : $$\frac{1}{2}S_{\infty} = \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}...$$ Use the formula for infinite GP in the RHS to get : $$\frac{1}{2}S_{\infty} = \frac{\frac{1}{2}}{1-\frac{1}{2}}$$ You will get $$S_{\infty}=2$$.

PS: In the first summation you can also use $$\cos hx$$ instead of $$\cos ix$$ because: $$\cos hx = \cos ix = \frac{e^{i.(ix)}+e^{-i.(ix)}}{2}=\frac{e^{x}+e^{-x}}{2}$$ Hope this helps .....

Observe that for any $$x \in \Bbb R$$ the Taylor series expansion of $$e^x$$ about $$x=0$$ is $$\sum_{n=0}^{\infty} \frac {x^n} {n!}.$$ Note that \begin{align} e+e^{-1} & = 2 \sum_{n=0}^{\infty} \frac {1} {(2n)!} \\ \implies \sum_{n=0}^{\infty} \frac {1} {(2n)!} & = \frac {e+e^{-1}} {2} =\cosh (1). \end{align}

For the second one observe that \begin{align} \sum_{n=0}^{\infty} \frac {n} {2^n} & = \sum_{n=1}^{\infty} \frac {n} {2^n} \\ & = \sum_{n=1}^{\infty} \frac {1} {2^n} + \sum_{n=2}^{\infty} \frac {1} {2^n} + \sum_{n=3}^{\infty} \frac {1} {2^n} + \cdots \\ & = 1 + \frac 1 2 + \frac {1} {2^2} + \cdots \\ & = \sum_{n=0}^{\infty} \frac {1} {2^n} \\ & = \frac {1} {1 - \frac 1 2} \\ & = 2. \end{align}